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Let the distribution function of the random variable $X$ be:$\qquad$ $F_X(x) = \begin{cases}0 \qquad if \qquad x < 0, \\ \frac{x}{3} \qquad if \qquad 0 \leq x < 1, \\ \frac{x}{2} \qquad if \qquad 1 \leq x < 2 \\ 1 \qquad if \qquad 2 \leq x \end{cases}$

Find the probability function $f_X(x)$

My try:

If we graph $F_X(x)$ we can tell it´s a "mixed" fucntion. Because it´s discontinuous at $x=1$ and it can be expresed as a sum of a discrete function $F^d_X(x)$ and a continous function $F^c_X(x)$:

$$F_X(x) = F^c_X(x) + F^d_X(x)$$ then $\qquad$ $F^d_X(x) = \begin{cases}0 \qquad if \qquad x \leq 1, \\ x \qquad if \qquad 1 < x \end{cases}$

and for $F^c_X(x)$ we got the equation $$F^c_X(x) = F_X(x) - F^d_X(x)$$ which give us the fucntion

$F^c_X(x) = \begin{cases}0 \qquad if \qquad x < 0, \\ \frac{x}{3} \qquad if \qquad 0 \leq x < 1, \\ \frac{x-2}{2} \qquad if \qquad 1 \leq x < 2 \\ 1 \qquad if \qquad 2 \leq x \end{cases}$

then $$f_X(x) = f^d_X(x) + f^c_X(x)$$ it would be easy to find.

I don´t know if this process is correct. Any help would be great. Thanks!

Sebas Martinez Santos
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1 Answers1

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There is not probability density function $f_X$ satisfying the distribution $F_X$. Note the relationship that $F_X(b) = \int_{-\infty}^b f_X(x)dx$ where we are using the Lebesgue integral. Then since $f_X$ is integrable, we must have that $F_X$ is continuous. See The Lebesgue integral of an integrable function is continuous.

Andrew Shedlock
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